Every generating isotone projection cone is latticial and correct

Abstract

?

Authors

G. Isac
Departement de Mathematiques, College Militaire Royal, St. Jean, Q&bee, Canada JOJ IR

A.B. Németh
Institutul de Matematica, Str. Republicii Nr. 37, 3400 Cluj-Napoca, Roumanie

Keywords

?

Paper coordinates

G. Isac, A.B. Nemeth, Every generating isotone projection cone is latticial and correct, J. Math. Anal. Appl., 147 (1990) no. 1, pp. 53-62
doi: 10.1016/0022-247X(90)90383-Q

PDF

About this paper

Journal

Journal of Mathematical Analysis and Applications

Publisher Name

Elsevier

Print ISSN

?

Online ISSN

?

Google Scholar Profile

?

1. G. P. BARKER, Perfect cones, Linear Alg Appl. 22 (1978) 211-221.
2. J. M. BORWEIN, D. T. YOST, Absolute norms on vector lattices, Proc. Edinburgh Math. Sm. 27 ( 1984) 215222.
3. B. IOCHUM, Cones autopolaires et algebres de Jordan, in “Lecture Notes in Math.,” Vol. 1049, Springer-Verlag, New York/Berlin, 1984.
4. G. ISAC, A. B. NEMETH, Monotonicity of metric projections onto positive cones of ordered euclidean spaces, Arch. Math. 46 (1986) 5688576; Corrigendum, Arch. Math. 49 (1987), 367-368.
5. G. ISAC, A. B. NEMETH, Isotone projection cones in Hilbert spaces and the complementarity problem, Boll. U.M.I. (7) 3-B (1989).
6. G. ISAC AND A. B. NEMETH, Ordered Hilbert spaces, preprint, 1987.
7. C. W. MTARTHUR, In what spaces is every closed normal cone regular? Proc. Edinburgh Math. Soc. (2) 17 (1970), 121-125.
8. J. MOREAU, Decomposition orthogonale d’un espace hilbertien selon deux cones mutuellement pollaires, C. R. Acad. Sci. Paris Ser Math. 225 (1962), 238-240.
9. F. RIESZ, Sur quelques notions fondamentales dans la theorie g&r&ale des operations lineaires, Mat. Term~szet~ l&es. 56 (1937), 145; Ann. qf Math. 41 (1940), 174-206.
10. E. H. ZARANTONELLO, Projections on convex sets in Hilbert space and spectral theory, in “Contributions to Nonlinear Functional Analysis” (E. H. Zarantonello, Ed.), pp. 237424, Academic Press, New York, 1971.
11. A. YOUDINE. Solution de deux problemes de la theorie des espaces semi-ordonnes. C. R. Acad. Sci. U.R.S.S. 27 (1939), 418-422.

Paper (preprint) in HTML form

1990-Nemeth-Every generating isotone projection cone is latticial

Every Generating Isotone Projection Cone Is Latticial and Correct

G. IsacDépartement de Mathématiques, Collège Militaire Royal, St. Jean, Québec, Canada J0J lR0

ANDA. B. NémethInstitutul de Matematica, Str. Republicii Nr. 37, 3400 Cluj-Napoca, RoumanieSubmitted by E. Stanley Lee

Received September 12, 1988

1. Introduction

In the last decade, ordered topological vector spaces have been studied a great deal, but ordered Hilbert spaces have mysteriously been neglected.
It is interesting to remark that the ordered Hilbert space has interesting and deep properties and important applications in areas such as Quantum Mechanics [3], Potential Theory, Analytic Manifold Theory, the study of positivity of Green's functions, Nonlinear Analysis, the study of the Complementarity Problem [6], etc.
In connection with the study of the Complementarity Problem we introduced in [5,4] the concept of isotone projection cone. Certainly, the isotone projection cones have applications in Numerical Analysis, Variational Inequalities, and Optimization by projection methods.
Let ( H , ( H , (H,(::)(H,\langle\rangle(H,, ) b e a H i l b e r t s p a c e a n d l e t K H ) b e a H i l b e r t s p a c e a n d l e t K H )beaHilbertspaceandletK sub H) be a Hilbert space and let K \subset H)beaHilbertspaceandletKH be a pointed convex cone; that is, (1) K + K K K + K K K+K sub KK+K \subset KK+KK, (2) ( λ R + ) ( λ K K ) λ R + ( λ K K ) (AA lambda inR_(+))(lambda K sub K)\left(\forall \lambda \in R_{+}\right)(\lambda K \subset K)(λR+)(λKK), and (3) K ( K ) = { 0 } K ( K ) = { 0 } K nn(-K)={0}K \cap(-K)=\{0\}K(K)={0}.
We denote by " <=\leqslant " the order defined on H H HHH by K K KKK, that is, x y y x K x y y x K x <= y<=>y-x in Kx \leqslant y \Leftrightarrow y-x \in KxyyxK.
If K K KKK is closed it is well known [10] that for every x H x H x in Hx \in HxH there exists a unique element P K ( x ) K P K ( x ) K P_(K)(x)in KP_{K}(x) \in KPK(x)K (called the projection of x x xxx onto K K KKK ) such that
x P K ( x ) x y , for every y K . x P K ( x ) x y ,  for every  y K . ||x-P_(K)(x)|| <= ||x-y||,quad" for every "y in K.\left\|x-P_{K}(x)\right\| \leqslant\|x-y\|, \quad \text { for every } y \in K .xPK(x)xy, for every yK.
The following problem has never previously been considered in the mathematical literature.
When is P K P K P_(K)P_{K}PK monotone increasing (isotone) with respect to the order defined by K K KKK ?
So, we say that K K KKK is an isotone projection if for every x , y H , x y x , y H , x y x,y in H,x <= yx, y \in H, x \leqslant yx,yH,xy implies that P K ( x ) P K ( y ) P K ( x ) P K ( y ) P_(K)(x) <= P_(K)(y)P_{K}(x) \leqslant P_{K}(y)PK(x)PK(y) [5,4].
We are interested in characterizing by a geometric or analytic property isotone projection cones.
In Euclidean space we know a characterization of isotone projection cones by a necessary and sufficient condition [4,5] but in infinite dimensional Hilbert spaces this problem is not solved.
In this note we prove that if K H K H K sub HK \subset HKH is a generating isotone projection cone in H H HHH, then it is latticial and correct.
But, if K H K H K sub HK \subset HKH is a closed convex cone, generating, latticial, and correct, is it an isotone projection? We do not know.

2. Definitions and Main Results

Let ( H , ( H , (H,(::)(H,\langle\rangle(H,, ) b e a H i l b e r t s p a c e . I f K H ) b e a H i l b e r t s p a c e . I f K H )beaHilbertspace.IfK sub H) be a Hilbert space. If K \subset H)beaHilbertspace.IfKH is a closed convex cone, then for every x H x H x in Hx \in HxH the projection P K ( x ) P K ( x ) P_(K)(x)P_{K}(x)PK(x) can be characterized by the relation
(1) x P K ( x ) , P K ( x ) y 0 ; y K ( see [10, Lemma 1.1] ) . (1) x P K ( x ) , P K ( x ) y 0 ; y K (  see [10, Lemma 1.1]  ) . {:(1)(:x-P_(K)(x),P_(K)(x)-y:) >= 0;quad AA y in K(" see [10, Lemma 1.1] ").:}\begin{equation*} \left\langle x-P_{K}(x), P_{K}(x)-y\right\rangle \geqslant 0 ; \quad \forall y \in K(\text { see [10, Lemma 1.1] }) . \tag{1} \end{equation*}(1)xPK(x),PK(x)y0;yK( see [10, Lemma 1.1] ).
The dual of K K KKK is K = { y H x , y 0 , x K } K = { y H x , y 0 , x K } K^(**)={y in H∣(:x,y:) >= 0,AA x in K}K^{*}=\{y \in H \mid\langle x, y\rangle \geqslant 0, \forall x \in K\}K={yHx,y0,xK}.
K K K^(**)K^{*}K is a closed convex cone and if K K KKK is generating, that is, if K K = H K K = H K-K=HK-K=HKK=H, then K K K^(**)K^{*}K is a pointed convex cone.
Besides (1) the theorem of Moreau will be the other basic instrument in our proof.
The polar of K K KKK is K = K K = K K^(@)=-K^(**)K^{\circ}=-K^{*}K=K. If K K KKK is closed then K = ( K ) K = K K=(K^(@))^(@)K=\left(K^{\circ}\right)^{\circ}K=(K).
If K K KKK and Q Q QQQ are two closed convex cones in H H HHH then we say that K K KKK and Q Q QQQ are mutually polar if K = Q K = Q K=Q^(@)K=Q^{\circ}K=Q (which implies K = Q K = Q K^(@)=QK^{\circ}=QK=Q ).
Theorem (Moreau [8]). If K K KKK and Q Q QQQ are two mutually polar convex cones in H H HHH and x , y , z H x , y , z H x,y,z in Hx, y, z \in Hx,y,zH then the following statements are equivalent:
(i) z = x + y , x K , y Q z = x + y , x K , y Q z=x+y,x in K,y in Qz=x+y, x \in K, y \in Qz=x+y,xK,yQ, and x , y = 0 x , y = 0 (:x,y:)=0\langle x, y\rangle=0x,y=0,
(ii) x = P K ( z ) x = P K ( z ) x=P_(K)(z)x=P_{K}(z)x=PK(z) and y = P Q ( z ) y = P Q ( z ) y=P_(Q)(z)y=P_{Q}(z)y=PQ(z).
Every convex cone will be supposed pointed.
Definition 1. A convex cone K H K H K sub HK \subset HKH is called an isotone projection if
(2) K is closed, (2) K  is closed,  {:(2)K" is closed, ":}\begin{equation*} K \text { is closed, } \tag{2} \end{equation*}(2)K is closed, 
whenever v u K v u K v-u in Kv-u \in KvuK it follows that P K ( v ) P K ( u ) K P K ( v ) P K ( u ) K P_(K)(v)-P_(K)(u)in KP_{K}(v)-P_{K}(u) \in KPK(v)PK(u)K.
Remark. If we consider on H H HHH the order " <=\leqslant " defined by K K KKK then condition (3) of Definition 1 says that P K P K P_(K)P_{K}PK is isotone with respect to the order defined by K K KKK, whence the reason for our terminology.
Observe that (3) in Definition 1 is a purely geometric definition and it does not depend on the position of K K KKK in H H HHH, while an order theoretic definition requires K K KKK to be a "cone with vertex at the origin," i.e., a cone in the sense of our above definition.
The space H H HHH ordered by K K KKK as above will be called a vector lattice and K K KKK will be called latticial if every pair of elements u u uuu and v v vvv in H H HHH has a least upper bound denoted by u v u v u vv vu \vee vuv.
Let K H K H K sub HK \subset HKH be a convex cone. A convex cone F K F K F sub KF \subset KFK is called a face of K K KKK if x K , y F x K , y F x in K,y in Fx \in K, y \in FxK,yF, and y x K y x K y-x in Ky-x \in KyxK imply x F x F x in Fx \in FxF.
(Note that this notion is less restrictive than the notion of face used in [10], which is in fact exposed face.)
Definition 2. The cone K K KKK in H H HHH will be called correct if for each of face F F FFF, we have
(4) P s p F ( K ) K , (4) P s p ¯ F ( K ) K , {:(4)P_( bar(sp)F)(K)sub K",":}\begin{equation*} P_{\overline{s p} F}(K) \subset K, \tag{4} \end{equation*}(4)PspF(K)K,
where s p F s p ¯ F bar(sp)F\overline{s p} FspF is the closed linear span of F F FFF.
Remark. The above notion of correct cone is related to that of perfect cone which turns out to coincide with correct cone for a self-dual cone (see Proposition 1 in [1]).
In [4] we gave a characterization of isotone projection cones in Euclidean spaces.
In [5] we considered problems concerning the facial structure of isotone projection cones as well their application to the Complementarity Problem.
This note concerns two necessary conditions for order to a cone K K KKK in the Hilbert space H H HHH to be an isotone projection. The main results can be summarized in the form:
THEOREM 1. If K K KKK is a generating isotone projection cone in H H HHH then it is latticial and correct.
It turns out (see Theorem 9 in Section 5) that our theorem above generalizes the necessary part of the main result in [4] (which is stated as Theorem 9 in Section 5).

3. Isotone Projection Cones Are Latticial

The cone K K KKK in H H HHH is called subdual if K K K K K subK^(**)K \subset K^{*}KK. We say that K K KKK is normal if there exists a constant γ > 0 γ > 0 gamma > 0\gamma>0γ>0 such that 0 x y 0 x y 0 <= x <= y0 \leqslant x \leqslant y0xy implies γ x y γ x y gamma||x|| <= ||y||\gamma\|x\| \leqslant\|y\|γxy.
Obviously if H H HHH is ordered by K K KKK and for every u , v H u , v H u,v in Hu, v \in Hu,vH satisfying 0 u v 0 u v 0 <= u <= v0 \leqslant u \leqslant v0uv we have that u v u v ||u|| <= ||v||\|u\| \leqslant\|v\|uv we deduce that K K KKK is normal.
The ordered space H H HHH and its positive cone K K KKK is called regular if every decreasing sequence of elements in K K KKK is convergent.
This implies in particular that each increasing order bounded sequence in H H HHH is convergent.
Proposition 2. If K K KKK is an isotone projection cone in the Hilbert space H H HHH then it is normal and regular.
Proof. In our paper [5] we proved (Theorem 2.2) that every isotone projection cone is subdual.
We prove now that K K KKK is normal. Indeed, suppose 0 u v 0 u v 0 <= u <= v0 \leqslant u \leqslant v0uv. Then v u K v u K v-u in Kv-u \in KvuK and since K K KKK is subdual, v u K v u K v-u inK^(**)v-u \in K^{*}vuK and we have v u , v 0 v u , v 0 (:v-u,v:) >= 0\langle v-u, v\rangle \geqslant 0vu,v0 and v u , u 0 v u , u 0 (:v-u,u:) >= 0\langle v-u, u\rangle \geqslant 0vu,u0 whence it follows that u 2 u , v v 2 u 2 u , v v 2 ||u||^(2) <= (:u,v:) <= ||v||^(2)\|u\|^{2} \leqslant\langle u, v\rangle \leqslant\|v\|^{2}u2u,vv2, which shows that K K KKK is normal.
A result of McArthur [7] shows that every closed normal cone in a Banach space is regular if this space does not contain any subspace isomorphic with c 0 c 0 c_(0)c_{0}c0 (the Banach space of all sequences of real numbers converging to zero equipped with the supremum norm). Since this holds for a Hilbert space we conclude that K K KKK is also regular.
Proposition 3. Every generating isotone projection cone K K KKK in the Hilbert space H H HHH is latticial.
Proof. We begin by proving the following auxiliary result.
Lemma 4. Let K K KKK be a closed generating cone in H H HHH.
If there exist a u + K , b v + K a u + K , b v + K a in u+K,b in v+Ka \in u+K, b \in v+Kau+K,bv+K with the properties a = P u + K ( b ) a = P u + K ( b ) a=P_(u+K)(b)a=P_{u+K}(b)a=Pu+K(b), b = P v + K ( a ) b = P v + K ( a ) b=P_(v+K)(a)b=P_{v+K}(a)b=Pv+K(a), then a = b ( u + K ) ( v + K ) a = b ( u + K ) ( v + K ) a=b in(u+K)nn(v+K)a=b \in(u+K) \cap(v+K)a=b(u+K)(v+K).
Proof of the Lemma. Since K K KKK is generating there exists an element w ( u + K ) ( v + K ) w ( u + K ) ( v + K ) w in(u+K)nn(v+K)w \in(u+K) \cap(v+K)w(u+K)(v+K). (This is equivalent with the assertion that there exists an element w w www with u w u w u <= wu \leqslant wuw and v w v w v <= wv \leqslant wvw.
Since K K KKK is generating, there exist u 1 , u 2 , v 1 , v 2 K u 1 , u 2 , v 1 , v 2 K u_(1),u_(2),v_(1),v_(2)in Ku_{1}, u_{2}, v_{1}, v_{2} \in Ku1,u2,v1,v2K such that u = u 1 u 2 u = u 1 u 2 u=u_(1)-u_(2)u=u_{1}-u_{2}u=u1u2 and v = v 1 v 2 v = v 1 v 2 v=v_(1)-v_(2)v=v_{1}-v_{2}v=v1v2.
Hence u u 1 , v v 1 u u 1 , v v 1 u <= u_(1),v <= v_(1)u \leqslant u_{1}, v \leqslant v_{1}uu1,vv1 and as u 1 u 1 u_(1)u_{1}u1 and v 1 v 1 v_(1)v_{1}v1 are in K K KKK it follows that u 1 u 1 + v 1 u 1 u 1 + v 1 u_(1) <= u_(1)+v_(1)u_{1} \leqslant u_{1}+v_{1}u1u1+v1 and v 1 u 1 + v 1 v 1 u 1 + v 1 v_(1) <= u_(1)+v_(1)v_{1} \leqslant u_{1}+v_{1}v1u1+v1. Thus we can take w = u 1 + v 1 w = u 1 + v 1 w=u_(1)+v_(1)w=u_{1}+v_{1}w=u1+v1 to have the required property.)
We have by the characterization (1) of the projections the relations
a P v + K ( a ) , P v + K ( a ) w 0 a P v + K ( a ) , P v + K ( a ) w 0 (:a-P_(v+K)(a),P_(v+K)(a)-w:) >= 0\left\langle a-P_{v+K}(a), P_{v+K}(a)-w\right\rangle \geqslant 0aPv+K(a),Pv+K(a)w0
and
b P u + K ( b ) , P u + K ( b ) w 0 . b P u + K ( b ) , P u + K ( b ) w 0 . (:b-P_(u+K)(b),P_(u+K)(b)-w:) >= 0.\left\langle b-P_{u+K}(b), P_{u+K}(b)-w\right\rangle \geqslant 0 .bPu+K(b),Pu+K(b)w0.
Using the conditions in the assertion of the lemma on a a aaa and b b bbb, the second relation can be written in the form P v + κ ( a ) a , a w 0 P v + κ ( a ) a , a w 0 (:P_(v+kappa)(a)-a,a-w:) >= 0\left\langle P_{v+\kappa}(a)-a, a-w\right\rangle \geqslant 0Pv+κ(a)a,aw0. We have on the other hand,
P v + K ( a ) a , a w = P v + K ( a ) a , ( a P v + K ( a ) ) + ( P v + K ( a ) w ) = ( P v + K ( a ) a 2 + a P v + K ( a ) , P v + K ( a ) w ) P v + K ( a ) a , a w = P v + K ( a ) a , a P v + K ( a ) + P v + K ( a ) w = P v + K ( a ) a 2 + a P v + K ( a ) , P v + K ( a ) w {:[(:P_(v+K)(a)-a,a-w:)=(:P_(v+K)(a)-a,(a-P_(v+K)(a))+(P_(v+K)(a)-w):)],[=-(||P_(v+K)(a)-a||^(2)+(:a-P_(v+K)(a),P_(v+K)(a)-w:))]:}\begin{aligned} \left\langle P_{v+K}(a)-a, a-w\right\rangle & =\left\langle P_{v+K}(a)-a,\left(a-P_{v+K}(a)\right)+\left(P_{v+K}(a)-w\right)\right\rangle \\ & =-\left(\left\|P_{v+K}(a)-a\right\|^{2}+\left\langle a-P_{v+K}(a), P_{v+K}(a)-w\right\rangle\right) \end{aligned}Pv+K(a)a,aw=Pv+K(a)a,(aPv+K(a))+(Pv+K(a)w)=(Pv+K(a)a2+aPv+K(a),Pv+K(a)w)
which together with the above deduced relations shows that
P v + K ( a ) a = b a = 0 P v + K ( a ) a = b a = 0 ||P_(v+K)(a)-a||=||b-a||=0\left\|P_{v+K}(a)-a\right\|=\|b-a\|=0Pv+K(a)a=ba=0
We now prove Proposition 3.
Let u u uuu and v v vvv be arbitrary elements in H H HHH. We shall show using the isotone projection property of K K KKK that they admit a least upper bound u v u v u vv vu \vee vuv by constructing effectively this element. If u u uuu and v v vvv are comparable we have nothing to do.
Suppose they are not comparable.
Let w w www be an arbitrary upper bound of the set { u , v } { u , v } {u,v}\{u, v\}{u,v}, i.e., an arbitrary element of ( u + K ) ( v + K ) ( u + K ) ( v + K ) (u+K)nn(v+K)(u+K) \cap(v+K)(u+K)(v+K) which is not empty since K K KKK is generating by hypothesis.
We see next that if P K P K P_(K)P_{K}PK is isotone, then P y + K P y + K P_(y+K)P_{y+K}Py+K is isotone too for an arbitrary y y yyy in H H HHH. This follows directly from the relation P y + K ( x ) = P K ( x y ) + y P y + K ( x ) = P K ( x y ) + y P_(y+K)(x)=P_(K)(x-y)+yP_{y+K}(x)= P_{K}(x-y)+yPy+K(x)=PK(xy)+y, which can be verified directly using (1). Hence P u + K P u + K P_(u+K)P_{u+K}Pu+K and P v + K P v + K P_(v+K)P_{v+K}Pv+K are both isotone.
Since neither of the convex sets u + K u + K u+Ku+Ku+K and v + K v + K v+Kv+Kv+K is contained in the other, using Lemma 4 we conclude that the relations u = P u + K ( v ) u = P u + K ( v ) u=P_(u+K)(v)u=P_{u+K}(v)u=Pu+K(v) and v = P v + K ( u ) v = P v + K ( u ) v=P_(v+K)(u)v=P_{v+K}(u)v=Pv+K(u) cannot hold simultaneously.
Suppose u P u + K ( v ) u + K u P u + K ( v ) u + K u!=P_(u+K)(v)in u+Ku \neq P_{u+K}(v) \in u+KuPu+K(v)u+K.
Hence u P u + K ( v ) P u + K ( w ) = w u P u + K ( v ) P u + K ( w ) = w u <= P_(u+K)(v) <= P_(u+K)(w)=wu \leqslant P_{u+K}(v) \leqslant P_{u+K}(w)=wuPu+K(v)Pu+K(w)=w, since P u + K P u + K P_(u+K)P_{u+K}Pu+K is isotone.
Let us consider the operators Q := P v + K P u + K Q := P v + K P u + K Q:=P_(v+K)@P_(u+K)Q:=P_{v+K} \circ P_{u+K}Q:=Pv+KPu+K and R := P u + K P v + K R := P u + K P v + K R:=P_(u+K)@P_(v+K)R:=P_{u+K} \circ P_{v+K}R:=Pu+KPv+K. They are isotone since P u + K P u + K P_(u+K)P_{u+K}Pu+K and P v + K P v + K P_(v+K)P_{v+K}Pv+K are. Put v n = Q n ( v ) , u 1 = P u + K ( v ) v n = Q n ( v ) , u 1 = P u + K ( v ) v_(n)=Q^(n)(v),u_(1)=P_(u+K)(v)v_{n}=Q^{n}(v), u_{1}=P_{u+K}(v)vn=Qn(v),u1=Pu+K(v), and u n = R n 1 ( u 1 ) u n = R n 1 u 1 u_(n)=R^(n-1)(u_(1))u_{n}=R^{n-1}\left(u_{1}\right)un=Rn1(u1). Then we have the relations
v v 1 v n w v v 1 v n w v <= v_(1) <= cdots <= v_(n) <= cdots <= wv \leqslant v_{1} \leqslant \cdots \leqslant v_{n} \leqslant \cdots \leqslant wvv1vnw
and
u u 1 u 2 u n w , u u 1 u 2 u n w , u <= u_(1) <= u_(2) <= cdots <= u_(n) <= cdots <= w,u \leqslant u_{1} \leqslant u_{2} \leqslant \cdots \leqslant u_{n} \leqslant \cdots \leqslant w,uu1u2unw,
since u u 1 , v v 1 , P u + K , Q u u 1 , v v 1 , P u + K , Q u <= u_(1),v <= v_(1),P_(u+K),Qu \leqslant u_{1}, v \leqslant v_{1}, P_{u+K}, Quu1,vv1,Pu+K,Q, and R R RRR are isotone and since P u + K ( w ) = Q ( w ) = R ( w ) = w P u + K ( w ) = Q ( w ) = R ( w ) = w P_(u+K)(w)=Q(w)=R(w)=wP_{u+K}(w)= Q(w)=R(w)=wPu+K(w)=Q(w)=R(w)=w.
(Obviously P v + K P u + K ( v ) v + K , P v + K P u + K ( v ) v + K , P_(v+K)@P_(u+K)(v)in v+K,quadP_{v+K} \circ P_{u+K}(v) \in v+K, \quadPv+KPu+K(v)v+K, hence v P v + K P u + K ( v ) = Q ( v ) = v 1 v P v + K P u + K ( v ) = Q ( v ) = v 1 quad v <= P_(v+K)@P_(u+K)(v)=Q(v)=v_(1)quad\quad v \leqslant P_{v+K} \circ P_{u+K}(v)= Q(v)=v_{1} \quadvPv+KPu+K(v)=Q(v)=v1 and u 1 = P u + K ( v ) P u + K P v + K P u + K ( v ) u 1 = P u + K ( v ) P u + K P v + K P u + K ( v ) quadu_(1)=P_(u+K)(v) <= P_(u+K)@P_(v+K)@P_(u+K)(v)\quad u_{1}=P_{u+K}(v) \leqslant P_{u+K} \circ P_{v+K} \circ P_{u+K}(v)u1=Pu+K(v)Pu+KPv+KPu+K(v), that is, u 1 R ( u 1 ) = u 2 u 1 R u 1 = u 2 u_(1) <= R(u_(1))=u_(2)u_{1} \leqslant R\left(u_{1}\right)=u_{2}u1R(u1)=u2, etc.)
We have further,
v n = Q n ( v ) = ( P v + K P u + K ) n ( v ) = P r + K ( P u + K P v + K ) n P u + K ( v ) (5) = P v + K R n 1 ( u 1 ) = P v + K ( u n ) v n = Q n ( v ) = P v + K P u + K n ( v ) = P r + K P u + K P v + K n P u + K ( v ) (5) = P v + K R n 1 u 1 = P v + K u n {:[v_(n)=Q^(n)(v)=(P_(v+K)P_(u+K))^(n)(v)=P_(r+K)(P_(u+K)P_(v+K))^(n)P_(u+K)(v)],[(5)=P_(v+K)R^(n*1)(u_(1))=P_(v+K)(u_(n))]:}\begin{align*} v_{n} & =Q^{n}(v)=\left(P_{v+K} P_{u+K}\right)^{n}(v)=P_{r+K}\left(P_{u+K} P_{v+K}\right)^{n} P_{u+K}(v) \\ & =P_{v+K} R^{n \cdot 1}\left(u_{1}\right)=P_{v+K}\left(u_{n}\right) \tag{5} \end{align*}vn=Qn(v)=(Pv+KPu+K)n(v)=Pr+K(Pu+KPv+K)nPu+K(v)(5)=Pv+KRn1(u1)=Pv+K(un)
and
(6) u n + 1 = R ( u n ) = P u + K P u + K ( u n ) = P u + K ( v n ) (6) u n + 1 = R u n = P u + K P u + K u n = P u + K v n {:(6)u_(n+1)=R(u_(n))=P_(u+K)@P_(u+K)(u_(n))=P_(u+K)(v_(n)):}\begin{equation*} u_{n+1}=R\left(u_{n}\right)=P_{u+K} \circ P_{u+K}\left(u_{n}\right)=P_{u+K}\left(v_{n}\right) \tag{6} \end{equation*}(6)un+1=R(un)=Pu+KPu+K(un)=Pu+K(vn)
Since the sequences { u n } u n {u_(n)}\left\{u_{n}\right\}{un} and { v n } v n {v_(n)}\left\{v_{n}\right\}{vn} are increasing and majorized by w w www, and since K K KKK is regular by Proposition 2 there exist the limits
(7) u 0 = lim u u n and v 0 = lim n v n (7) u 0 = lim u u n  and  v 0 = lim n v n {:(7)u_(0)=lim_(u rarr oo)u_(n)quad" and "quadv_(0)=lim_(n rarr oo)v_(n):}\begin{equation*} u_{0}=\lim _{u \rightarrow \infty} u_{n} \quad \text { and } \quad v_{0}=\lim _{n \rightarrow \infty} v_{n} \tag{7} \end{equation*}(7)u0=limuun and v0=limnvn
such that
(8) u u 0 w and v v 0 w , (8) u u 0 w  and  v v 0 w , {:(8)u <= u_(0) <= w quad" and "quad v <= v_(0) <= w",":}\begin{equation*} u \leqslant u_{0} \leqslant w \quad \text { and } \quad v \leqslant v_{0} \leqslant w, \tag{8} \end{equation*}(8)uu0w and vv0w,
since u u n w u u n w u <= u_(n) <= wu \leqslant u_{n} \leqslant wuunw and v v n w v v n w v <= v_(n) <= wv \leqslant v_{n} \leqslant wvvnw for every n n nnn and since K K KKK is closed.
From the continuity of projections the relations (5), (6), and (7) imply v 0 = P v + K ( u 0 ) v 0 = P v + K u 0 v_(0)=P_(v+K)(u_(0))v_{0}=P_{v+K}\left(u_{0}\right)v0=Pv+K(u0) and u 0 = P u + K ( v 0 ) u 0 = P u + K v 0 u_(0)=P_(u+K)(v_(0))u_{0}=P_{u+K}\left(v_{0}\right)u0=Pu+K(v0).
Using Lemma 4 again we deduce that u 0 = v 0 ( u + K ) ( v + K ) u 0 = v 0 ( u + K ) ( v + K ) u_(0)=v_(0)in(u+K)nn(v+K)u_{0}=v_{0} \in(u+K) \cap(v+K)u0=v0(u+K)(v+K). Since the upper bound w w www is arbitrary, from the relations (8) we obtain that in fact u 0 = v 0 = u v u 0 = v 0 = u v u_(0)=v_(0)=u vv vu_{0}=v_{0}=u \vee vu0=v0=uv and the proposition is proved.
We remark that the sequences { u n } u n {u_(n)}\left\{u_{n}\right\}{un} and { v n } v n {v_(n)}\left\{v_{n}\right\}{vn} can be stationary from a certain point onwards. (This can occur only when P u + K ( v ) v + K P u + K ( v ) v + K P_(u+K)(v)in v+KP_{u+K}(v) \in v+KPu+K(v)v+K.) But this case is also included in the above schema of the proof.

4. Isotone Projection Cones Are Correct

Proposition 5. For every face F F FFF of the isotone projection cone K K KKK in H H HHH the space s p F s p ¯ F bar(sp)F\overline{s p} FspF projects by P K P K P_(K)P_{K}PK onto F ¯ F ¯ bar(F)\bar{F}F¯ and F ¯ F ¯ bar(F)\bar{F}F¯ is an isotone projection cone in s p F s p ¯ F bar(sp)F\overline{s p} FspF.
Proof. Suppose that F F FFF is a face of K K KKK and consider z s p F z s p ¯ F z in bar(sp)Fz \in \overline{s p} FzspF. Then z = x y z = x y z=x-yz=x-yz=xy with x , y F K x , y F K x,y in F sub Kx, y \in F \subset Kx,yFK, whence z x z x z <= xz \leqslant xzx.
Since P K P K P_(K)P_{K}PK is isotone, one has 0 P K ( z ) P K ( x ) = x F 0 P K ( z ) P K ( x ) = x F 0 <= P_(K)(z) <= P_(K)(x)=x in F0 \leqslant P_{K}(z) \leqslant P_{K}(x)=x \in F0PK(z)PK(x)=xF, hence P K ( z ) F P K ( z ) F P_(K)(z)in FP_{K}(z) \in FPK(z)F.
Consider now z s p ^ F z s p ^ F z in widehat(sp)Fz \in \widehat{s p} Fzsp^F. Then there exists a sequence { z n } z n {z_(n)}\left\{z_{n}\right\}{zn} with z n s p ^ F z n s p ^ F z_(n)in widehat(sp)Fz_{n} \in \widehat{s p} Fznsp^F such that lim n z n = z lim n z n = z lim_(n rarr oo)z_(n)=z\lim _{n \rightarrow \infty} z_{n}=zlimnzn=z.
We have shown in the above paragraph that P K ( z n ) F P K z n F P_(K)(z_(n))in FP_{K}\left(z_{n}\right) \in FPK(zn)F for each n n nnn, hence P K ( z n ) P K ( z ) F ¯ P K z n P K ( z ) F ¯ P_(K)(z_(n))rarrP_(K)(z)in bar(F)P_{K}\left(z_{n}\right) \rightarrow P_{K}(z) \in \bar{F}PK(zn)PK(z)F¯ by continuity of P K P K P_(K)P_{K}PK [10]. This concludes at once that s p F s p ¯ F bar(sp)F\overline{s p} FspF projects onto F ¯ F ¯ bar(F)\bar{F}F¯ by P K P K P_(K)P_{K}PK, that P K s p F = P F ¯ s p F P K s p ¯ F = P F ¯ s p ¯ F P_(K∣ bar(sp)F)=P_( bar(F)∣ bar(sp)F)P_{K \mid \overline{s p} F}=P_{\bar{F} \mid \overline{s p} F}PKspF=PF¯spF, and that P F ¯ P F ¯ P_( bar(F))P_{\bar{F}}PF¯ is an isotone projection with respect to the order induced by F ¯ F ¯ bar(F)\bar{F}F¯ in s p F s p ¯ F bar(sp)F\overline{s p} FspF.
Remark. This proposition constitutes a generalization of Proposition 5.1 in [5].
Proposition 6. Every isotone projection cone in H H HHH is correct.
Proof. Assume the contrary. Then there exists a face F F FFF of K K KKK and an clement k k kkk of K K KKK such that z := P s p F ( k ) K z := P s p ¯ F ( k ) K z:=P_( bar(sp)F)(k)!in Kz:=P_{\overline{s p} F}(k) \notin Kz:=PspF(k)K.
Denote z 0 := P K ( z ) z 0 := P K ( z ) z_(0):=P_(K)(z)z_{0}:=P_{K}(z)z0:=PK(z). Since z S p F z S p ¯ F z in bar(Sp)Fz \in \overline{S p} FzSpF it follows by Proposition 5 that z 0 F ¯ z 0 F ¯ z_(0)in bar(F)z_{0} \in \bar{F}z0F¯.
We shall show first that a real number t ( 0 , 1 ) t ( 0 , 1 ) t in(0,1)t \in(0,1)t(0,1) can be determined such that for w := t k + ( 1 t ) z 0 w := t k + ( 1 t ) z 0 w:=tk+(1-t)z_(0)w:=t k+(1-t) z_{0}w:=tk+(1t)z0 one has
(9) z w , k z 0 = 0 . (9) z w , k z 0 = 0 . {:(9)(:z-w,k-z_(0):)=0.:}\begin{equation*} \left\langle z-w, k-z_{0}\right\rangle=0 . \tag{9} \end{equation*}(9)zw,kz0=0.
Indeed, we have
z t k ( 1 t ) z 0 , k z 0 = z k + ( 1 t ) ( k z 0 ) , k z 0 = z k , k z 0 + ( 1 t ) z 0 k 2 = z k , k z + z z 0 + ( 1 t ) z 0 k 2 = z k 2 + ( 1 t ) z 0 k 2 z t k ( 1 t ) z 0 , k z 0 = z k + ( 1 t ) k z 0 , k z 0 = z k , k z 0 + ( 1 t ) z 0 k 2 = z k , k z + z z 0 + ( 1 t ) z 0 k 2 = z k 2 + ( 1 t ) z 0 k 2 {:[(:z-tk-(1-t)z_(0),k-z_(0):)=(:z-k+(1-t)(k-z_(0)),k-z_(0):)],[=(:z-k,k-z_(0):)+(1-t)||z_(0)-k||^(2)],[=(:z-k,k-z+z-z_(0):)+(1-t)||z_(0)-k||^(2)],[=-||z-k||^(2)+(1-t)||z_(0)-k||^(2)]:}\begin{aligned} \left\langle z-t k-(1-t) z_{0}, k-z_{0}\right\rangle & =\left\langle z-k+(1-t)\left(k-z_{0}\right), k-z_{0}\right\rangle \\ & =\left\langle z-k, k-z_{0}\right\rangle+(1-t)\left\|z_{0}-k\right\|^{2} \\ & =\left\langle z-k, k-z+z-z_{0}\right\rangle+(1-t)\left\|z_{0}-k\right\|^{2} \\ & =-\|z-k\|^{2}+(1-t)\left\|z_{0}-k\right\|^{2} \end{aligned}ztk(1t)z0,kz0=zk+(1t)(kz0),kz0=zk,kz0+(1t)z0k2=zk,kz+zz0+(1t)z0k2=zk2+(1t)z0k2
since z k , z z 0 = 0 ( z z 0 s p F z k , z z 0 = 0 z z 0 s p ¯ F (:z-k,z-z_(0):)=0(z-z_(0)in bar(sp)F:}\left\langle z-k, z-z_{0}\right\rangle=0\left(z-z_{0} \in \overline{s p} F\right.zk,zz0=0(zz0spF and z k = P s p F ( k ) k z k = P s p ¯ F ( k ) k z-k=P_( bar(sp)F)(k)-kz-k=P_{\overline{s p} F}(k)-kzk=PspF(k)k is orthogonal to s p F s p ¯ F bar(sp)F\overline{s p} FspF ).
Since z k < z 0 k z k < z 0 k ||z-k|| < ||z_(0)-k||\|z-k\|<\left\|z_{0}-k\right\|zk<z0k by the definition of z z zzz, it follows that putting 1 t = z k 2 / z 0 k 2 < 1 1 t = z k 2 / z 0 k 2 < 1 1-t=||z-k||^(2)//||z_(0)-k||^(2) < 11-t=\|z-k\|^{2} /\left\|z_{0}-k\right\|^{2}<11t=zk2/z0k2<1, relation (9) holds. Obviously, w K w K w in Kw \in KwK.
From the definition of z 0 z 0 z_(0)z_{0}z0 we have using relation (1) that
(10) z z 0 , z 0 k = z P K ( z ) , P K ( z ) k 0 . (10) z z 0 , z 0 k = z P K ( z ) , P K ( z ) k 0 . {:(10)(:z-z_(0),z_(0)-k:)=(:z-P_(K)(z),P_(K)(z)-k:) >= 0.:}\begin{equation*} \left\langle z-z_{0}, z_{0}-k\right\rangle=\left\langle z-P_{K}(z), P_{K}(z)-k\right\rangle \geqslant 0 . \tag{10} \end{equation*}(10)zz0,z0k=zPK(z),PK(z)k0.
On the other hand, by the definition of w w www it follows that
z z 0 , z 0 k = z w + w z 0 , z 0 k = w z 0 , z 0 k = t k + ( 1 t ) z 0 z 0 , z 0 k = t k z 0 , z 0 k < 0 z z 0 , z 0 k = z w + w z 0 , z 0 k = w z 0 , z 0 k = t k + ( 1 t ) z 0 z 0 , z 0 k = t k z 0 , z 0 k < 0 {:[(:z-z_(0),z_(0)-k:)=(:z-w+w-z_(0),z_(0)-k:)=(:w-z_(0),z_(0)-k:)],[=(:tk+(1-t)z_(0)-z_(0),z_(0)-k:)=t(:k-z_(0),z_(0)-k:) < 0]:}\begin{aligned} \left\langle z-z_{0}, z_{0}-k\right\rangle & =\left\langle z-w+w-z_{0}, z_{0}-k\right\rangle=\left\langle w-z_{0}, z_{0}-k\right\rangle \\ & =\left\langle t k+(1-t) z_{0}-z_{0}, z_{0}-k\right\rangle=t\left\langle k-z_{0}, z_{0}-k\right\rangle<0 \end{aligned}zz0,z0k=zw+wz0,z0k=wz0,z0k=tk+(1t)z0z0,z0k=tkz0,z0k<0
This relation contradicts (10) and shows that our hypothesis that K K KKK is not correct is false.
So Theorem 1 is completely proved.

5. The Finite Dimensional Case

One dimensional faces of the cone K K KKK in Euclidean space R n R n R^(n)R^{n}Rn are called extreme rays.
The closed and generating cone K K KKK in R n R n R^(n)R^{n}Rn was called thin in [4] if
u , v 0 u , v 0 (:u,v:) <= 0\langle u, v\rangle \leqslant 0u,v0 for any two nonzero vectors u u uuu and v v vvv on any two different extreme rays of K 0 K 0 K^(0)K^{0}K0.
The basic result in [4] is the following.
Theorem 7. The generating cone K K KKK in R n R n R^(n)R^{n}Rn is an isotone projection if and only if it is thin.
How does this theorem relate to our Theorem 1?
To answer this question we prove first the following result.
Proposition 8. The closed generating cone K K KKK in the Euclidean space R n R n R^(n)R^{n}Rn is thin if and only if it is latticial and correct.
Proof. The latticiality of a thin cone follows from (i) in Lemma 2 of [4] and the characterization of Youdine of latticial cones in R n R n R^(n)R^{n}Rn [11].
From (iii) of the same lemma we have that the thin cone K K KKK and its polar K 0 K 0 K^(0)K^{0}K0 can be represented in the form K = cone { e 1 , e 2 , , e n } , K 0 = K = cone e 1 , e 2 , , e n , K 0 = K=cone{e_(1),e_(2),dots,e_(n)},K^(0)=K=\operatorname{cone}\left\{e_{1}, e_{2}, \ldots, e_{n}\right\}, K^{0}=K=cone{e1,e2,,en},K0= cone { u 1 , u 2 , , u n } u 1 , u 2 , , u n {u_(1),u_(2),dots,u_(n)}\left\{u_{1}, u_{2}, \ldots, u_{n}\right\}{u1,u2,,un} where conc M M MMM is the conic hull of M , e i , u j , i , j = 1 , , n M , e i , u j , i , j = 1 , , n M,e_(i),u_(j),i,j=1,dots,nM, e_{i}, u_{j}, i, j= 1, \ldots, nM,ei,uj,i,j=1,,n, are unit vectors, and e i , u j = 0 , u i , u j 0 e i , u j = 0 , u i , u j 0 (:e_(i),u_(j):)=0,(:u_(i),u_(j):) <= 0\left\langle e_{i}, u_{j}\right\rangle=0,\left\langle u_{i}, u_{j}\right\rangle \leqslant 0ei,uj=0,ui,uj0 if i j , i , j = 1 , 2 , , n i j , i , j = 1 , 2 , , n i!=j,i,j=1,2,dots,ni \neq j, i, j= 1,2, \ldots, nij,i,j=1,2,,n.
Let P P PPP denote the projection onto Π ( u n ) := { x R n x , u n = 0 } = s p { e 1 , e 2 , , e n 1 } Π u n := x R n x , u n = 0 = s p e 1 , e 2 , , e n 1 Pi(u_(n)):={x inR^(n)∣(:x,u_(n):)=0}=sp{e_(1),e_(2),dots,e_(n-1)}\Pi\left(u_{n}\right):=\left\{x \in R^{n} \mid\left\langle x, u_{n}\right\rangle=0\right\}= s p\left\{e_{1}, e_{2}, \ldots, e_{n-1}\right\}Π(un):={xRnx,un=0}=sp{e1,e2,,en1}.
We show that P ( K ) K P ( K ) K P(K)sub KP(K) \subset KP(K)K. To this end we have to show only that P ( e n ) K P e n K P(e_(n))in KP\left(e_{n}\right) \in KP(en)K, since an arbitrary element x x xxx of K K KKK is of the form x = c 1 e 1 + + c n 1 e n 1 + c n e n x = c 1 e 1 + + c n 1 e n 1 + c n e n x=c_(1)e_(1)+cdots+c_(n-1)e_(n-1)+c_(n)e_(n)quadx=c_{1} e_{1}+\cdots+ c_{n-1} e_{n-1}+c_{n} e_{n} \quadx=c1e1++cn1en1+cnen with c i 0 , i = 1 , 2 , , n , c i 0 , i = 1 , 2 , , n , quadc_(i) >= 0,quad i=1,2,dots,n,quad\quad c_{i} \geqslant 0, \quad i=1,2, \ldots, n, \quadci0,i=1,2,,n, and P ( x ) = c 1 e 1 + + c n 1 e n 1 + c n P ( e n ) P ( x ) = c 1 e 1 + + c n 1 e n 1 + c n P e n quad P(x)=c_(1)e_(1)+cdots+c_(n-1)e_(n-1)+c_(n)P(e_(n))\quad P(x)=c_{1} e_{1}+\cdots+ c_{n-1} e_{n-1}+c_{n} P\left(e_{n}\right)P(x)=c1e1++cn1en1+cnP(en).
But since u n u n u_(n)u_{n}un is of unit length we have
P ( e n ) = e n e n , u n u n P e n = e n e n , u n u n P(e_(n))=e_(n)-(:e_(n),u_(n):)u_(n)P\left(e_{n}\right)=e_{n}-\left\langle e_{n}, u_{n}\right\rangle u_{n}P(en)=enen,unun
and
P ( e n ) , u j = e n , u n u n , u j 0 for j = 1 , , n 1 P e n , u j = e n , u n u n , u j 0  for  j = 1 , , n 1 (:P(e_(n)),u_(j):)=-(:e_(n),u_(n):)(:u_(n),u_(j):) <= 0quad" for "quad j=1,dots,n-1\left\langle P\left(e_{n}\right), u_{j}\right\rangle=-\left\langle e_{n}, u_{n}\right\rangle\left\langle u_{n}, u_{j}\right\rangle \leqslant 0 \quad \text { for } \quad j=1, \ldots, n-1P(en),uj=en,unun,uj0 for j=1,,n1
(since e n , u n < 0 e n , u n < 0 (:e_(n),u_(n):) < 0\left\langle e_{n}, u_{n}\right\rangle<0en,un<0 and u n , u j 0 u n , u j 0 (:u_(n),u_(j):) <= 0\left\langle u_{n}, u_{j}\right\rangle \leqslant 0un,uj0 by hypothesis), while P ( e n ) , u n = 0 P e n , u n = 0 (:P(e_(n)),u_(n):)=0\left\langle P\left(e_{n}\right), u_{n}\right\rangle=0P(en),un=0. This is enough to see that P ( e n ) ( K 0 ) 0 = K P e n K 0 0 = K P(e_(n))in(K^(0))^(0)=KP\left(e_{n}\right) \in\left(K^{0}\right)^{0}=KP(en)(K0)0=K. Apply induction to sec that K K KKK is correct.
Suppose now that K K KKK is latticial and correct and show that it is thin.
It is immediate that K K KKK is closed and generating. Since K K KKK is latticial, K 0 K 0 K^(0)K^{0}K0 is latticial too [9], and hence it is a cone hull of n n nnn unit vectors u 1 , u 2 , , u n u 1 , u 2 , , u n u_(1),u_(2),dots,u_(n)u_{1}, u_{2}, \ldots, u_{n}u1,u2,,un which generate all the extreme rays of K 0 K 0 K^(0)K^{0}K0 (this follows from the theorem of Youdine cited above).
Assume that K K KKK is not thin. Then there exists at least a pair of vectors in the set { u 1 , u 2 , , u n } u 1 , u 2 , , u n {u_(1),u_(2),dots,u_(n)}\left\{u_{1}, u_{2}, \ldots, u_{n}\right\}{u1,u2,,un} which form an acute angle. Suppose for instance that u 1 , u 2 > 0 u 1 , u 2 > 0 (:u_(1),u_(2):) > 0\left\langle u_{1}, u_{2}\right\rangle>0u1,u2>0.
The faces F i := { x K x , u i = 0 } , i = 1 , 2 F i := x K x , u i = 0 , i = 1 , 2 F_(i):={x in K∣(:x,u_(i):)=0},i=1,2F_{i}:=\left\{x \in K \mid\left\langle x, u_{i}\right\rangle=0\right\}, i=1,2Fi:={xKx,ui=0},i=1,2, are different ( n 1 ) ( n 1 ) (n-1)(n-1)(n1)-dimensional faces of K K KKK.
Their spans are Π ( u i ) := { x H x , u i = 0 } , i = 1 , 2 Π u i := x H x , u i = 0 , i = 1 , 2 Pi(u_(i)):={x in H∣(:x,u_(i):)=0},i=1,2\Pi\left(u_{i}\right):=\left\{x \in H \mid\left\langle x, u_{i}\right\rangle=0\right\}, i=1,2Π(ui):={xHx,ui=0},i=1,2. Pick x F 1 F 2 x F 1 F 2 x inF_(1)\\F_(2)x \in F_{1} \backslash F_{2}xF1F2 and project it into Π ( u 2 ) Π u 2 Pi(u_(2))\Pi\left(u_{2}\right)Π(u2). Since u 1 u 1 u_(1)u_{1}u1 is of unit norm by hypothesis, the projection w w www of x x xxx into Π ( u 1 ) Π u 1 Pi(u_(1))\Pi\left(u_{1}\right)Π(u1) is of form w = x x , u 1 u 1 w = x x , u 1 u 1 w=x-(:x,u_(1):)u_(1)w=x-\left\langle x, u_{1}\right\rangle u_{1}w=xx,u1u1.
Observe first that since
(11) K = { y R n y , u 0 , i = 1 , 2 , , n } (11) K = y R n y , u 0 , i = 1 , 2 , , n {:(11)K={y inR^(n)∣(:y,u:) <= 0,i=1,2,dots,n}:}\begin{equation*} K=\left\{y \in R^{n} \mid\langle y, u\rangle \leqslant 0, i=1,2, \ldots, n\right\} \tag{11} \end{equation*}(11)K={yRny,u0,i=1,2,,n}
and from the definition of x x xxx, we must have x , u 2 = 0 , x , u 1 < 0 x , u 2 = 0 , x , u 1 < 0 (:x,u_(2):)=0,(:x,u_(1):) < 0\left\langle x, u_{2}\right\rangle=0,\left\langle x, u_{1}\right\rangle<0x,u2=0,x,u1<0.
Hence we have w , u 2 = x , u 2 x , u 1 u 1 , u 2 = x , u 1 u 1 , u 2 > 0 w , u 2 = x , u 2 x , u 1 u 1 , u 2 = x , u 1 u 1 , u 2 > 0 (:w,u_(2):)=(:x,u_(2):)-(:x,u_(1):)(:u_(1),u_(2):)=-(:x,u_(1):)(:u_(1),u_(2):) > 0\left\langle w, u_{2}\right\rangle=\left\langle x, u_{2}\right\rangle-\left\langle x, u_{1}\right\rangle\left\langle u_{1}, u_{2}\right\rangle=-\left\langle x, u_{1}\right\rangle \left\langle u_{1}, u_{2}\right\rangle>0w,u2=x,u2x,u1u1,u2=x,u1u1,u2>0.
This relation shows by (11) that w K w K w!in Kw \notin KwK, that is, P s p F 1 ( K ) K P s p ¯ F 1 ( K ) K P_( bar(sp)F_(1))(K)⊄KP_{\overline{s p} F_{1}}(K) \not \subset KPspF1(K)K and hence K K KKK cannot be correct.
Proposition 8 allows us to give the following equivalent formulation of Theorem 7.
Theorem 9. The generating cone K K KKK in R n R n R^(n)R^{n}Rn is an isotone projection if and only if it is latticial and correct.
Remarks. (1) We have in the proof of Proposition 3 a very interesting constructive method of defining the element sup ( u , v ) ( u v ) sup ( u , v ) ( u v ) s u p(u,v)(u vv v)\sup (u, v)(u \vee v)sup(u,v)(uv) where u , v H u , v H u,v in Hu, v \in Hu,vH and H H HHH is ordered by an isotone projection cone.
Given an ordered Hilbert space ( H , , , ) ( H , , , ) (H,(::),, <= )(H,\langle\rangle,, \leqslant)(H,,,) where the order " <=\leqslant " is defined by an isotone projection cone, then for every u , v H u , v H u,v in Hu, v \in Hu,vH (if u u uuu and v v vvv are not comparable) then the element u v u v u vv vu \vee vuv is the limit of a sequence obtained by an iterative method. (See the proof of Proposition 3.)
(2) We recall [6,2] that a Hilbert lattice is a Hilbert space ( H , H , H,(:H,\langleH,, ) o r d e r e d b y a c l o s e d p o i n t e d c o n v e x c o n e K ) o r d e r e d b y a c l o s e d p o i n t e d c o n v e x c o n e K :))orderedbyaclosedpointedconvexconeK\rangle ) ordered by a closed pointed convex cone K)orderedbyaclosedpointedconvexconeK such that: (1) H H HHH is a vector lattice; (2) x y x y ||x|| <= ||y||\|x\| \leqslant\|y\|xy, whenever 0 x y 0 x y 0 <= x <= y0 \leqslant x \leqslant y0xy; and (3) for every x H x H x in Hx \in HxH, | x | = x | x | = x |||x|||=||x||\||x|\|=\|x\||x|=x (that is, the norm is absolute).
We note that an interesting study of absolute norms can be found in [2].
In our paper [5] we proved that if K H K H K sub HK \subset HKH is a self-dual cone (that is, K = K ) K = K {:K=K^(**))\left.K=K^{*}\right)K=K), then K K KKK is an isotone projection if and only if ( H , H , H,(:H,\langleH,, ) i s a ) i s a :))isa\rangle ) is a)isa Hilbert lattice with respect to the order defined by K K KKK.
From Theorem 1 and Proposition 2 we deduce that if ( H , ( H , (H,(::)(H,\langle\rangle(H,, ) i s a ) i s a )isa) is a)isa Hilbert space ordered by an isotone projection cone K K KKK such that K K K K K!=K^(**)K \neq K^{*}KK then the norm ||||\|\| of H H HHH is not absolute.
We finish this note with the following problem:
In an infinite dimensional Hilbert space, is a generating, latticial, and correct cone also an isotone projection cone?

References

  1. G. P. Barker, Perfect cones, Linear Algehra Appl. 22 (1978), 211221.
  2. J. M. Borwein and D. T. Yost, Absolute norms on vector lattices, Proc. Edinburgh Math. Soc. 27 (1984), 215-222.
  3. B. Iochum, Cônes autopolaires et algèbres de Jordan, in "Lecture Notes in Math.," Vol. 1049, Springer-Verlag, New York/Berlin, 1984.
  4. G. Isac and A. B. Németh, Monotonicity of metric projections onto positive cones of ordered euclidean spaces, Arch. Math. 46 (1986), 568-576; Corrigendum, Arch. Mat. 49 (1987), 367-368.
  5. G. Isac and A. B. Németh, Isotone projection cones in Hilbert spaces and the complementarity problem, Boll. U.M.I. (7) 3-B (1989).
  6. G. Isac and A. B. Németh, Ordered Hilbert spaces, preprint, 1987.
  7. C. W. McArthur, In what spaces is every closed normal cone regular? Proc. Edinhurgh Math. Soc. (2) 17 (1970), 121-125.
  8. J. Moreau, Décomposition orthogonale d'un espace hilbertien selon deux cônes mutuellement pollaires, C. R. Acad. Sci. Paris Sér. I Math. 225 (1962), 238-240.
  9. F. Riesz, Sur quelques notions fondamentales dans la théorie générale des opérations linéaires, Mat. Természett Értes. 56 (1937), 1-45; Ann. of Math. 41 (1940), 174-206.
  10. E. H. Zarantonello, Projections on convex sets in Hilbert space and spectral theory, in "Contributions to Nonlinear Functional Analysis" (E. H. Zarantonello, Ed.), pp. 237-424, Academic Press, New York, 1971.
  11. A. Youdine, Solution de deux problèmes de la théorie des espaces semi-ordonnés, C. R. Acad. Sci. U.R.S.S. 27 (1939), 418-422.
1990

Related Posts